3.27.76 \(\int \frac {(-1+2 k^2) x-2 k^4 x^3+k^4 x^5}{((1-x^2) (1-k^2 x^2))^{2/3} (1-d+(d-2 k^2) x^2+k^4 x^4)} \, dx\) [2676]

Optimal. Leaf size=241 \[ -\frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}{2 \sqrt [3]{d}-2 \sqrt [3]{d} x^2+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}\right )}{2 \sqrt [3]{d}}-\frac {\log \left (-\sqrt [3]{d}+\sqrt [3]{d} x^2+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 \sqrt [3]{d}}+\frac {\log \left (d^{2/3}-2 d^{2/3} x^2+d^{2/3} x^4+\left (\sqrt [3]{d}-\sqrt [3]{d} x^2\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{4/3}\right )}{4 \sqrt [3]{d}} \]

[Out]

-1/2*3^(1/2)*arctan(3^(1/2)*(1+(-k^2-1)*x^2+k^2*x^4)^(2/3)/(2*d^(1/3)-2*d^(1/3)*x^2+(1+(-k^2-1)*x^2+k^2*x^4)^(
2/3)))/d^(1/3)-1/2*ln(-d^(1/3)+d^(1/3)*x^2+(1+(-k^2-1)*x^2+k^2*x^4)^(2/3))/d^(1/3)+1/4*ln(d^(2/3)-2*d^(2/3)*x^
2+d^(2/3)*x^4+(d^(1/3)-d^(1/3)*x^2)*(1+(-k^2-1)*x^2+k^2*x^4)^(2/3)+(1+(-k^2-1)*x^2+k^2*x^4)^(4/3))/d^(1/3)

________________________________________________________________________________________

Rubi [A]
time = 150.14, antiderivative size = 1, normalized size of antiderivative = 0.00, number of steps used = 993, number of rules used = 5, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {1608, 1976, 6847, 6820, 8} \begin {gather*} 0 \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[((-1 + 2*k^2)*x - 2*k^4*x^3 + k^4*x^5)/(((1 - x^2)*(1 - k^2*x^2))^(2/3)*(1 - d + (d - 2*k^2)*x^2 + k^4*x^4
)),x]

[Out]

0

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1976

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))*((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Int[u*(a*c*e + (b*c
+ a*d)*e*x^n + b*d*e*x^(2*n))^p, x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

\begin {align*} \int \frac {\left (-1+2 k^2\right ) x-2 k^4 x^3+k^4 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (d-2 k^2\right ) x^2+k^4 x^4\right )} \, dx &=\int \frac {x \left (-1+2 k^2-2 k^4 x^2+k^4 x^4\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (d-2 k^2\right ) x^2+k^4 x^4\right )} \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {-1+2 k^2-2 k^4 x+k^4 x^2}{\left ((1-x) \left (1-k^2 x\right )\right )^{2/3} \left (1-d+\left (d-2 k^2\right ) x+k^4 x^2\right )} \, dx,x,x^2\right )\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {-1+2 k^2-2 k^4 x+k^4 x^2}{(1-x)^{2/3} \left (1-k^2 x\right )^{2/3} \left (1-d+\left (d-2 k^2\right ) x+k^4 x^2\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{1-k^2 x} \left (-1+2 k^2-k^2 x\right )}{(1-x)^{2/3} \left (1-d+\left (d-2 k^2\right ) x+k^4 x^2\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \text {Subst}\left (\int \left (\frac {\left (-k^2+\frac {k^2 \sqrt {d-4 k^2+4 k^4}}{\sqrt {d}}\right ) \sqrt [3]{1-k^2 x}}{(1-x)^{2/3} \left (d-2 k^2-\sqrt {d} \sqrt {d-4 k^2+4 k^4}+2 k^4 x\right )}+\frac {\left (-k^2-\frac {k^2 \sqrt {d-4 k^2+4 k^4}}{\sqrt {d}}\right ) \sqrt [3]{1-k^2 x}}{(1-x)^{2/3} \left (d-2 k^2+\sqrt {d} \sqrt {d-4 k^2+4 k^4}+2 k^4 x\right )}\right ) \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=-\frac {\left (k^2 \left (1-\frac {\sqrt {d-4 k^2+4 k^4}}{\sqrt {d}}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{1-k^2 x}}{(1-x)^{2/3} \left (d-2 k^2-\sqrt {d} \sqrt {d-4 k^2+4 k^4}+2 k^4 x\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}-\frac {\left (k^2 \left (1+\frac {\sqrt {d-4 k^2+4 k^4}}{\sqrt {d}}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{1-k^2 x}}{(1-x)^{2/3} \left (d-2 k^2+\sqrt {d} \sqrt {d-4 k^2+4 k^4}+2 k^4 x\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=-\frac {\left (k^2 \left (1-\frac {\sqrt {d-4 k^2+4 k^4}}{\sqrt {d}}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )\right ) \text {Subst}\left (\int \frac {\sqrt [3]{-\frac {1}{-1+k^2}+\frac {k^2 x}{-1+k^2}}}{(1-x)^{2/3} \left (d-2 k^2-\sqrt {d} \sqrt {d-4 k^2+4 k^4}+2 k^4 x\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \sqrt [3]{\frac {-1+k^2 x^2}{-1+k^2}}}-\frac {\left (k^2 \left (1+\frac {\sqrt {d-4 k^2+4 k^4}}{\sqrt {d}}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )\right ) \text {Subst}\left (\int \frac {\sqrt [3]{-\frac {1}{-1+k^2}+\frac {k^2 x}{-1+k^2}}}{(1-x)^{2/3} \left (d-2 k^2+\sqrt {d} \sqrt {d-4 k^2+4 k^4}+2 k^4 x\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \sqrt [3]{\frac {-1+k^2 x^2}{-1+k^2}}}\\ &=\frac {3 k^2 \left (1-\frac {\sqrt {d-4 k^2+4 k^4}}{\sqrt {d}}\right ) \left (1-x^2\right ) \left (1-k^2 x^2\right ) F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 k^4 \left (1-x^2\right )}{d-2 k^2 \left (1-k^2\right )-\sqrt {d} \sqrt {d-4 k^2+4 k^4}}\right )}{2 \left (d-2 k^2 \left (1-k^2\right )-\sqrt {d} \sqrt {d-4 k^2+4 k^4}\right ) \sqrt [3]{\frac {1-k^2 x^2}{1-k^2}} \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {3 k^2 \left (\sqrt {d}+\sqrt {d-4 k^2+4 k^4}\right ) \left (1-x^2\right ) \left (1-k^2 x^2\right ) F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 k^4 \left (1-x^2\right )}{d-2 k^2 \left (1-k^2\right )+\sqrt {d} \sqrt {d-4 k^2+4 k^4}}\right )}{2 \sqrt {d} \left (d-2 k^2 \left (1-k^2\right )+\sqrt {d} \sqrt {d-4 k^2+4 k^4}\right ) \sqrt [3]{\frac {1-k^2 x^2}{1-k^2}} \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 14.23, size = 201, normalized size = 0.83 \begin {gather*} \frac {\left (-1+x^2\right )^{2/3} \left (-1+k^2 x^2\right )^{2/3} \left (-2 \sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \left (-1+k^2 x^2\right )^{2/3}}{-2 \sqrt [3]{d} \sqrt [3]{-1+x^2}+\left (-1+k^2 x^2\right )^{2/3}}\right )-2 \log \left (\sqrt [3]{d} \sqrt [3]{-1+x^2}+\left (-1+k^2 x^2\right )^{2/3}\right )+\log \left (d^{2/3} \left (-1+x^2\right )^{2/3}-\sqrt [3]{d} \sqrt [3]{-1+x^2} \left (-1+k^2 x^2\right )^{2/3}+\left (-1+k^2 x^2\right )^{4/3}\right )\right )}{4 \sqrt [3]{d} \left (\left (-1+x^2\right ) \left (-1+k^2 x^2\right )\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1 + 2*k^2)*x - 2*k^4*x^3 + k^4*x^5)/(((1 - x^2)*(1 - k^2*x^2))^(2/3)*(1 - d + (d - 2*k^2)*x^2 + k
^4*x^4)),x]

[Out]

((-1 + x^2)^(2/3)*(-1 + k^2*x^2)^(2/3)*(-2*Sqrt[3]*ArcTan[(Sqrt[3]*(-1 + k^2*x^2)^(2/3))/(-2*d^(1/3)*(-1 + x^2
)^(1/3) + (-1 + k^2*x^2)^(2/3))] - 2*Log[d^(1/3)*(-1 + x^2)^(1/3) + (-1 + k^2*x^2)^(2/3)] + Log[d^(2/3)*(-1 +
x^2)^(2/3) - d^(1/3)*(-1 + x^2)^(1/3)*(-1 + k^2*x^2)^(2/3) + (-1 + k^2*x^2)^(4/3)]))/(4*d^(1/3)*((-1 + x^2)*(-
1 + k^2*x^2))^(2/3))

________________________________________________________________________________________

Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (2 k^{2}-1\right ) x -2 k^{4} x^{3}+k^{4} x^{5}}{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )^{\frac {2}{3}} \left (1-d +\left (-2 k^{2}+d \right ) x^{2}+k^{4} x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*k^2-1)*x-2*k^4*x^3+k^4*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(1-d+(-2*k^2+d)*x^2+k^4*x^4),x)

[Out]

int(((2*k^2-1)*x-2*k^4*x^3+k^4*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(1-d+(-2*k^2+d)*x^2+k^4*x^4),x)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*k^2-1)*x-2*k^4*x^3+k^4*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(1-d+(-2*k^2+d)*x^2+k^4*x^4),x, algori
thm="maxima")

[Out]

integrate((k^4*x^5 - 2*k^4*x^3 + (2*k^2 - 1)*x)/((k^4*x^4 - (2*k^2 - d)*x^2 - d + 1)*((k^2*x^2 - 1)*(x^2 - 1))
^(2/3)), x)

________________________________________________________________________________________

Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*k^2-1)*x-2*k^4*x^3+k^4*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(1-d+(-2*k^2+d)*x^2+k^4*x^4),x, algori
thm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*k**2-1)*x-2*k**4*x**3+k**4*x**5)/((-x**2+1)*(-k**2*x**2+1))**(2/3)/(1-d+(-2*k**2+d)*x**2+k**4*x*
*4),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*k^2-1)*x-2*k^4*x^3+k^4*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(1-d+(-2*k^2+d)*x^2+k^4*x^4),x, algori
thm="giac")

[Out]

integrate((k^4*x^5 - 2*k^4*x^3 + (2*k^2 - 1)*x)/((k^4*x^4 - (2*k^2 - d)*x^2 - d + 1)*((k^2*x^2 - 1)*(x^2 - 1))
^(2/3)), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {k^4\,x^5-2\,k^4\,x^3+x\,\left (2\,k^2-1\right )}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{2/3}\,\left (k^4\,x^4-d+x^2\,\left (d-2\,k^2\right )+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((k^4*x^5 - 2*k^4*x^3 + x*(2*k^2 - 1))/(((x^2 - 1)*(k^2*x^2 - 1))^(2/3)*(k^4*x^4 - d + x^2*(d - 2*k^2) + 1)
),x)

[Out]

int((k^4*x^5 - 2*k^4*x^3 + x*(2*k^2 - 1))/(((x^2 - 1)*(k^2*x^2 - 1))^(2/3)*(k^4*x^4 - d + x^2*(d - 2*k^2) + 1)
), x)

________________________________________________________________________________________